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Find sum of all unique elements in the array for K queries

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  • Last Updated : 18 Nov, 2021

Given an arrays arr[] in which initially all elements are 0 and another array Q[][] containing K queries where every query represents a range [L, R], the task is to add 1 to each subarrays where each subarray is defined by the the range [L, R], and return sum of all unique elements.
Note: One-based indexing is used in the Q[][] array to signify the ranges.
Examples: 
 

Input: arr[] = { 0, 0, 0, 0, 0, 0 }, Q[][2] = {{1, 3}, {4, 6}, {3, 4}, {3, 3}} 
Output:
Explanation: 
Initially the array is arr[] = { 0, 0, 0, 0, 0, 0 }. 
Query 1: arr[] = { 1, 1, 1, 0, 0, 0 }. 
Query 2: arr[] = { 1, 1, 1, 1, 1, 1 }. 
Query 3: arr[] = { 1, 1, 2, 2, 1, 1 }. 
Query 4: arr[] = { 1, 1, 3, 2, 1, 1 }. 
Hence unique elements are {1, 3, 2}. Thus sum = 1 + 3 + 2 = 6.
Input: arr[] = { 0, 0, 0, 0, 0, 0, 0, 0 }, Q[][2] = {{1, 4}, {5, 5}, {7, 8}, {8, 8}} 
Output:
Explanation: 
Initially the array is arr[] = { 0, 0, 0, 0, 0, 0, 0, 0 }. 
After processing all queries, arr[] = { 1, 1, 1, 1, 1, 0, 1, 2 }. 
Hence unique elements are {1, 0, 2}. Thus sum = 1 + 0 + 2 = 3. 
 

 

Approach: The idea is to increment the value by 1 and decrement the array by 1 at L and R+1 index respectively for processing each query. Then, Compute the prefix sum of the array to find the final array after Q queries. As explained in this article. Finally, compute the sum of the unique elements with the help of the hash-map.
Below is the implementation of the above approach: 
 

C++




// C++ implementation to find the
// sum of all unique elements of
// the array after Q queries
 
#include <bits/stdc++.h>
 
using namespace std;
 
 
// Function to find the sum of
// unique elements after Q Query
int uniqueSum(int A[], int R[][2],
            int N, int M)
{
    // Updating the array after
    // processing each query
    for (int i = 0; i < M; ++i) {
 
        int l = R[i][0], r = R[i][1] + 1;
 
        // Making it to 0-indexing
        l--;
        r--;
        A[l]++;
 
        if (r < N)
            A[r]--;
    }
 
    // Iterating over the array
    // to get the final array
    for (int i = 1; i < N; ++i) {
        A[i] += A[i - 1];
    }
 
    // Variable to store the sum
    int ans = 0;
 
    // Hash to maintain previously
    // occured elements
    unordered_set<int> s;
 
    // Loop to find the maximum sum
    for (int i = 0; i < N; ++i) {
 
        if (s.find(A[i]) == s.end())
            ans += A[i];
 
        s.insert(A[i]);
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int A[] = { 0, 0, 0, 0, 0, 0 };
    int R[][2]
        = { { 1, 3 }, { 4, 6 },
            { 3, 4 }, { 3, 3 } };
 
    int N = sizeof(A) / sizeof(A[0]);
    int M = sizeof(R) / sizeof(R[0]);
 
    cout << uniqueSum(A, R, N, M);
 
    return 0;
}

Java




// Java implementation to find the
// sum of all unique elements of
// the array after Q queries
 
import java.util.*;
 
class GFG{
 
// Function to find the sum of
// unique elements after Q Query
static int uniqueSum(int A[], int R[][],
                     int N, int M)
{
    // Updating the array after
    // processing each query
    for (int i = 0; i < M; ++i)
    {
        int l = R[i][0], r = R[i][1] + 1;
 
        // Making it to 0-indexing
        l--;
        r--;
        A[l]++;
         
        if (r < N)
            A[r]--;
    }
 
    // Iterating over the array
    // to get the final array
    for (int i = 1; i < N; ++i)
    {
        A[i] += A[i - 1];
    }
 
    // Variable to store the sum
    int ans = 0;
 
    // Hash to maintain previously
    // occured elements
    HashSet<Integer> s = new HashSet<Integer>();
 
    // Loop to find the maximum sum
    for (int i = 0; i < N; ++i)
    {
        if (!s.contains(A[i]))
            ans += A[i];
 
        s.add(A[i]);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int A[] = { 0, 0, 0, 0, 0, 0 };
    int R[][] = { { 1, 3 }, { 4, 6 },
                  { 3, 4 }, { 3, 3 } };
    int N = A.length;
    int M = R.length;
    System.out.print(uniqueSum(A, R, N, M));
}
}
 
// This code is contributed by gauravrajput1

Python 3




# Python implementation to find the
# sum of all unique elements of
# the array after Q queries
 
# Function to find the sum of
# unique elements after Q Query
def uniqueSum(A, R, N, M) :
 
    # Updating the array after
    # processing each query
    for i in range(0, M) :
 
        l = R[i][0]
        r = R[i][1] + 1
 
        # Making it to 0-indexing
        l -= 1
        r -= 1
        A[l] += 1
 
        if (r < N) :
            A[r] -= 1
 
    # Iterating over the array
    # to get the final array
    for i in range(1, N) :
        A[i] += A[i - 1]
 
    # Variable to store the sum
    ans = 0
 
    # Hash to maintain previously
    # occured elements
    s = {chr}
 
    # Loop to find the maximum sum
    for i in range(0, N) :
        if (A[i] not in s) :
            ans += A[i]
 
        s.add(A[i])
 
    return ans
 
# Driver code
A = [ 0, 0, 0, 0, 0, 0 ]
R = [ [ 1, 3 ], [ 4, 6 ],
      [ 3, 4 ], [ 3, 3 ] ]
N = len(A)
M = len(R)
 
print(uniqueSum(A, R, N, M))
 
# This code is contributed by Sanjit_Prasad

C#




// C# implementation to find the
// sum of all unique elements of
// the array after Q queries
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the sum of
// unique elements after Q Query
static int uniqueSum(int []A, int [,]R,
                     int N, int M)
{
     
    // Updating the array after
    // processing each query
    for(int i = 0; i < M; ++i)
    {
       int l = R[i, 0], r = R[i, 1] + 1;
        
       // Making it to 0-indexing
       l--;
       r--;
       A[l]++;
        
       if (r < N)
           A[r]--;
    }
 
    // Iterating over the array
    // to get the readonly array
    for(int i = 1; i < N; ++i)
    {
       A[i] += A[i - 1];
    }
 
    // Variable to store the sum
    int ans = 0;
 
    // Hash to maintain previously
    // occured elements
    HashSet<int> s = new HashSet<int>();
 
    // Loop to find the maximum sum
    for(int i = 0; i < N; ++i)
    {
       if (!s.Contains(A[i]))
           ans += A[i];
       s.Add(A[i]);
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int []A = { 0, 0, 0, 0, 0, 0 };
    int [,]R = { { 1, 3 }, { 4, 6 },
                 { 3, 4 }, { 3, 3 } };
                  
    int N = A.Length;
    int M = R.GetLength(0);
     
    Console.Write(uniqueSum(A, R, N, M));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
// Javascript implementation to find the
// sum of all unique elements of
// the array after Q queries
 
// Function to find the sum of
// unique elements after Q Query
function uniqueSum(A, R, N, M)
{
 
    // Updating the array after
    // processing each query
    for (let i = 0; i < M; ++i)
    {
        let l = R[i][0], r = R[i][1] + 1;
   
        // Making it to 0-indexing
        l--;
        r--;
        A[l]++;
           
        if (r < N)
            A[r]--;
    }
   
    // Iterating over the array
    // to get the final array
    for (let i = 1; i < N; ++i)
    {
        A[i] += A[i - 1];
    }
   
    // Variable to store the sum
    let ans = 0;
   
    // Hash to maintain previously
    // occured elements
    let s = new Set();
   
    // Loop to find the maximum sum
    for (let i = 0; i < N; ++i)
    {
        if (!s.has(A[i]))
            ans += A[i];
   
        s.add(A[i]);
    }
    return ans;
}
 
// Driver code
      let A = [ 0, 0, 0, 0, 0, 0 ];
    let R = [[ 1, 3 ], [ 4, 6 ],
                  [ 3, 4 ], [ 3, 3 ]];
    let N = A.length;
    let M = R.length;
    document.write(uniqueSum(A, R, N, M));
 
// This code is contributed by code_hunt.
</script>

Output: 

6

 

Time Complexity: O(M + N)

Auxiliary Space: O(N)


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